The Hamming distance between two integers is the number of positions at which the corresponding bits are different.
Given an integer array nums
, return the sum of Hamming distances between all the pairs of the integers in nums
.
Example 1:
Input: nums = [4,14,2] Output: 6 Explanation: In binary representation, the 4 is 0100, 14 is 1110, and 2 is 0010 (just showing the four bits relevant in this case). The answer will be: HammingDistance(4, 14) + HammingDistance(4, 2) + HammingDistance(14, 2) = 2 + 2 + 2 = 6.
Example 2:
Input: nums = [4,14,4] Output: 4
Constraints:
1 <= nums.length <= 104
0 <= nums[i] <= 109
- The answer for the given input will fit in a 32-bit integer.
class Solution:
def totalHammingDistance(self, nums: List[int]) -> int:
ans = 0
for i in range(31):
a = b = 0
for v in nums:
t = (v >> i) & 1
if t:
a += 1
else:
b += 1
ans += a * b
return ans
class Solution {
public int totalHammingDistance(int[] nums) {
int ans = 0;
for (int i = 0; i < 31; ++i) {
int a = 0, b = 0;
for (int v : nums) {
int t = (v >> i) & 1;
a += t;
b += t ^ 1;
}
ans += a * b;
}
return ans;
}
}
class Solution {
public:
int totalHammingDistance(vector<int>& nums) {
int ans = 0;
for (int i = 0; i < 31; ++i) {
int a = 0, b = 0;
for (int& v : nums) {
int t = (v >> i) & 1;
a += t;
b += t ^ 1;
}
ans += a * b;
}
return ans;
}
};
func totalHammingDistance(nums []int) int {
ans := 0
for i := 0; i < 31; i++ {
a, b := 0, 0
for _, v := range nums {
t := (v >> i) & 1
a += t
b += t ^ 1
}
ans += a * b
}
return ans
}